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Ex 9.3
Ex 9.3, 2
Ex 9.3, 3
Ex 9.3, 4 Important
Ex 9.3, 5
Ex 9.3, 6
Ex 9.3, 7 Important
Ex 9.3, 8
Ex 9.3, 9 Important
Ex 9.3, 10 Important
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13
Ex 9.3, 14
Ex 9.3, 15 Important
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Ex 9.3, 17 Important
Ex 9.3, 18
Ex 9.3, 19 Important
Ex 9.3, 20 Important
Ex 9.3, 21
Ex 9.3, 22 Important
Ex 9.3, 23 (MCQ)
Last updated at May 29, 2023 by Teachoo
Ex 9.3, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑑𝑦/𝑑𝑥=(1 − cos𝑥)/(1 + cos𝑥 ) 𝑑𝑦/𝑑𝑥 = (1 − cos𝑥)/(1 + cos𝑥 ) We know that cos 2x = 2cos2 x − 1 Putting x = 𝑥/2 cos 2𝑥/2 = 2 cos2 𝑥/2 − 1 cos x = 2 cos2 𝑥/2 − 1 1 + cos x = 2cos2 𝑥/2 We know cos 2x = 1 − 2sin2 x Putting x = 𝑥/2 cos2 2𝑥/2 = 1 − 2 sin2 𝑥/2 cos x = 1 − 2sin2 𝑥/2 1 − cos x = 2sin2 𝑥/2 Putting values in equation 𝒅𝒚/𝒅𝒙 = (𝟐 〖〖𝐬𝐢𝐧〗^𝟐 〗〖𝒙/𝟐〗 )/(𝟐 〖〖𝐜𝐨𝐬〗^𝟐 〗〖𝒙/𝟐〗 ) 𝑑𝑦 = (〖sin^2 〗〖𝑥/2〗 )/cos^(2 )〖𝑥/2〗 𝑑𝑥 𝑑𝑦 = tan2 𝑥/2 𝑑𝑥 Putting tan2 𝑥/2 = sec2 𝑥/2 − 1 𝒅𝒚 = ("sec2 " 𝒙/𝟐 " − 1" )𝒅𝒙 Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x − 1 Putting x = 𝑥/2 tan2 𝑥/2 = sec2 𝑥/2 − 1 ∫1▒𝒅𝒚 = ∫1▒(𝒔𝒆𝒄𝟐 𝒙/𝟐−𝟏) 𝒅𝒙 y = ∫1▒〖𝑠𝑒𝑐2 𝑥/2 𝑑𝑥− ∫1▒𝑑𝑥〗 y = 𝟏/(𝟏/𝟐) tan 𝒙/𝟐 − x + C y = 𝟐 tan 𝒙/𝟐 − x + C is the general solution (As ∫1▒〖𝑠𝑒𝑐2 (𝑎𝑥) 𝑑𝑥〗 = 1/𝑎 tan ax + C)